Problem: Simplify and expand the following expression: $ \dfrac{4}{5q + 15}- \dfrac{2}{4q + 8}+ \dfrac{5q}{q^2 + 5q + 6} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{4}{5q + 15} = \dfrac{4}{5(q + 3)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{2}{4q + 8} = \dfrac{2}{4(q + 2)}$ We can factor the quadratic in the third term: $ \dfrac{5q}{q^2 + 5q + 6} = \dfrac{5q}{(q + 3)(q + 2)}$ Now we have: $ \dfrac{4}{5(q + 3)}- \dfrac{2}{4(q + 2)}+ \dfrac{5q}{(q + 3)(q + 2)} $ The least common multiple of the denominators is: $ 20(q + 3)(q + 2)$ In order to get the first term over $20(q + 3)(q + 2)$ , multiply by $\dfrac{4(q + 2)}{4(q + 2)}$ $ \dfrac{4}{5(q + 3)} \times \dfrac{4(q + 2)}{4(q + 2)} = \dfrac{16(q + 2)}{20(q + 3)(q + 2)} $ In order to get the second term over $20(q + 3)(q + 2)$ , multiply by $\dfrac{5(q + 3)}{5(q + 3)}$ $ \dfrac{2}{4(q + 2)} \times \dfrac{5(q + 3)}{5(q + 3)} = \dfrac{10(q + 3)}{20(q + 3)(q + 2)} $ In order to get the third term over $20(q + 3)(q + 2)$ , multiply by $\dfrac{20}{20}$ $ \dfrac{5q}{(q + 3)(q + 2)} \times \dfrac{20}{20} = \dfrac{100q}{20(q + 3)(q + 2)} $ Now we have: $ \dfrac{16(q + 2)}{20(q + 3)(q + 2)} - \dfrac{10(q + 3)}{20(q + 3)(q + 2)} + \dfrac{100q}{20(q + 3)(q + 2)} $ $ = \dfrac{ 16(q + 2) - 10(q + 3) + 100q} {20(q + 3)(q + 2)} $ Expand: $ = \dfrac{16q + 32 - 10q - 30 + 100q}{20q^2 + 100q + 120} $ $ = \dfrac{106q + 2}{20q^2 + 100q + 120}$ Simplify: $ = \dfrac{53q + 1}{10q^2 + 50q + 60}$